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For one toss of a certain coin, the probability that the out

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For one toss of a certain coin, the probability that the out [#permalink]

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New post 09 Jul 2012, 07:50
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For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5
[Reveal] Spoiler: OA

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Re: For one toss of a certain coin, the probability that the out [#permalink]

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New post 09 Jul 2012, 07:57
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Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5


\(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: \(P(h=5)=0.6^5\);

4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\).

Answer: E.
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Re: For one toss of a certain coin, the probability that the out [#permalink]

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New post 19 Sep 2014, 20:00
Bunuel wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5


\(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: \(P(h=5)=0.6^5\);

4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\).

Answer: E.


Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.
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Re: For one toss of a certain coin, the probability that the out [#permalink]

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New post 20 Sep 2014, 13:22
kpali wrote:
Bunuel wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5


\(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: \(P(h=5)=0.6^5\);

4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\).

Answer: E.


Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.


Think logically, the case of 4 heads and 1 tail can occur in several ways (each of which having the same probability), so we have to account for the order here.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: For one toss of a certain coin, the probability that the out [#permalink]

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For one toss of a certain coin, the probability that the out [#permalink]

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New post 31 May 2017, 23:17
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

1. Probability of getting 4 heads and one tail is 5(0.6)^4 * (0.4). 5 is nothing but 5C4 ways. of 4 heads happening.
2. Probability of getting heads all the 5 times is (0.6)^5
3. Total probability is answer E.
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Re: For one toss of a certain coin, the probability that the out [#permalink]

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New post 25 Jun 2017, 12:29
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Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5



Refer to the solution in the picture
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Solution Head.jpeg [ 48.14 KiB | Viewed 763 times ]


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Re: For one toss of a certain coin, the probability that the out [#permalink]

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New post 08 Aug 2017, 07:46
mihir0710 wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5



Refer to the solution in the picture


Thanks for the lucid explanation..
In GMAT can we expect problems related to Bayes’ theorem?
Please suggest...
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Re: For one toss of a certain coin, the probability that the out   [#permalink] 08 Aug 2017, 07:46
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